How do you multiply #(x+3)(x+4) #?

1 Answer
Aug 11, 2015

Answer:

#(x+3) * (x+4) = x^2 + 7*x + 12#

Explanation:

The rigorous but rather lengthy explanation involves the distributive law of multiplication of a sum of two numbers by the third one and commutative law for both operations, addition and multiplication.

The distributive law of multiplication states:
#a*(b+c)=a*b+a*c#

The commutative law of addition states:
#a+b = b+a#

The commutative law of multiplication states"
#a*b = b*a#

All the above laws are used (usually without explicit mentioning) when performing the operations of this problem.

Using the distributive law described above for
#a=x+3#, #b=x# and #c=4#,
we obtain:
#(x+3) * (x+4) = (x+3) * x + (x+3) * 4#

Now let's use the commutative law of multiplication to rewrite it as
#(x+3) * (x+4) = x * (x+3) + 4 * (x+3)#

Now use the distributive law twice, first for
#a=x#, #b=x# and #c=3#
obtaining
#x * (x+3) = x*x + x*3#
and another time for
#a=4#, #b=x# and #c=3#
obtaining
#4 * (x+3) = 4*x + 4*3#

Let's use these equalities in the original expression:
#(x+3) * (x+4) = x*x + x*3 + 4*x + 4*3#

Further simplification involves the following transformations:
#x*x = x^2# (just for convenience)
Using commutative and distributive laws,
#x*3+4*x = x*3+x*4=x*(3+4)=x*7=7*x#
Obviously,
#4*3=12#

Our expression looks now as follows:
#(x+3) * (x+4) = x^2 + 7*x + 12#
This is the final representation of the original expression as a polynomial of the second degree.

Of course, most students do all these transformations without even thinking about the laws they are based upon and without writing an intermediary results. But these skills develop as a result of numerous exercises. Practice makes it perfect.